Cantor Set is Nowhere Dense
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Theorem
Let $\left({\mathcal C, \tau_d}\right)$ be the Cantor set considered as a topological subspace of the real number space $\R$ under the Euclidean topology $\tau_d$.
Then $\mathcal C$ is nowhere dense in $\left[{0 \,.\,.\, 1}\right]$.
Proof 1
From Cantor Set Closed in Real Number Space, $\mathcal C$ is closed.
So from Closed Set Equals its Closure:
- $\mathcal C^- = \mathcal C$
where $\mathcal C^-$ denotes the closure of $\mathcal C$.
Let $0 \le a < b \le 1$.
Then $I = \left({a \,.\,.\, b}\right)$ is an open interval of $\left[{0 \,.\,.\, 1}\right]$.
Let $\epsilon = b - a$.
Clearly $\epsilon > 0$.
Let $n \in \N$ such that $3^{-n} < \epsilon$.
Need to fill in the details to show that between $a$ and $b$ there is an interval $3^{-\left({n+1}\right)}$ wide that contains numbers with $1$ in its $n+1$th digit in its ternary expansion (or something like that) - intuitively obvious but fiddly to prove.
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So there exists an open interval of $\left[{0 \,.\,.\, 1}\right]$ which has been deleted from $\left[{0 \,.\,.\, 1}\right]$ during the process of creating $\mathcal C$.
Thus no open interval of $\left[{0 \,.\,.\, 1}\right]$ is disjoint from all the open intervals deleted from $\left[{0 \,.\,.\, 1}\right]$.
So any open interval of $\left[{0 \,.\,.\, 1}\right]$ can not be a subset of $\mathcal C = \mathcal C^-$.
Hence the result, by definition of nowhere dense.
$\blacksquare$
Proof 2
Let $\mathcal C_n$ denote the set $C_{n-1}$ with the middle open intervals of length $\frac{1}{3^{n}}$ removed from every one of the $2^{n-1}$ closed intervals, where $\mathcal C_0 = [0,1]$, and $\mathcal C_\infty = \mathcal C$.
Then the length of every interval in $C_n$ is $\frac{1}{3^n} = 3^{-n}$.
Let $0 \le a < b \le 1$.
Then $(a,b)\subset [0,1]$ is an open interval.
Let $n \in \N$ such that $3^{-n} < b-a$.
Then the length of every interval in $C_n$ is $3^{-n}<b-a$.
Therefore no interval of length $b-a$ exists in $\mathcal C_n$.
Therefore no interval of length $b-a$ exists in $\mathcal C = \mathcal C_\infty \subset \mathcal C_n$.
Since the interval $(a,b)$ was of arbitrary length, there do not exist any open intervals in $\mathcal C$.
Hence the result, by definition of nowhere dense.
$\blacksquare$
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Sources
- Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (1970)... (previous)... (next): $\text{II}: \ 29: \ 4$
- René L. Schilling: Measures, Integrals and Martingales (2005)... (previous)... (next) $\S 7$: Problem $10 \ \text{(v)}$
